Preimages All Unique iff Injection

Theorem

Let $f: S \to T$ be a mapping.

Let $f^{-1}$ be the inverse of $f$.

Let $f^{-1} \left({t}\right)$ be the preimage of $t \in T$.

Or we can consider the preimage of the subset $\left\{{t}\right\} \subseteq T$.

Then $f^{-1} \left({t}\right) = f^{-1} \left({\left\{{t}\right\}}\right)$ is a singleton for all $t \in T$ iff $f$ is an injection.

Proof

Follows immediately from the definition of injection.

• Let $\exists t \in T: s_1, s_2 \in f^{-1} \left({t}\right), s_1 \ne s_2$.

Then we have:

$f \left({s_1}\right) = f \left({s_2}\right)$ but $s_1 \ne s_2$

and so $f$ is not an injection.

So, by the Rule of Transposition, if $f$ is an injection then $f^{-1} \left({t}\right)$ has no more than one element.

• Suppose $f$ is not an injection.

Then by definition:

$\exists s_1, s_2 \in S, s_1 \ne s_2: f \left({s_1}\right) = f \left({s_2}\right) = t$

By definition of preimage of $t \in T$:

$s_1 \in f^{-1} \left({t}\right), s_2 \in f^{-1} \left({t}\right)$

and so: $\left\{{s_1, s_2}\right\} \subseteq f^{-1} \left({t}\right)$

So if $f$ is not an injection then $f^{-1} \left({t}\right)$ has more than one element for at least one $t \in T$.

Hence the result.

$\blacksquare$

From the definition of the image of a singleton, we have that $f^{-1} \left({t}\right) = f^{-1} \left({\left\{{t}\right\}}\right)$ and the result also can be seen to hold for $f^{-1} \left({\left\{{t}\right\}}\right)$.

Sources

• Paul R. Halmos: Naive Set Theory (1960)... (previous)... (next): $\S 10$: Inverses and Composites
• H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability (1996): Appendix $\text{A}.5$: Proposition $\text{A}.5.1: 1 \ \text{(b)}$